1000 Solved Problems in Modern Physics

3.3 Solutions 173

3.21 u=Ce−kr
∫∞
0 |u|

(^2) dr=c 2 ∫∞
0 e
− 2 kr=c^2
2 k=^1
C=

√

2 k

The probability that the neutron – proton separation in the deuteron exceeds

Ris

P=

∫∞

R

|u|^2 dr= 2 k

∫∞

R

e−^2 krdr

=e−^2 kR=e−(2×^0.^232 ×2)≈ 0. 4

Average distance of interaction

<r>=

∫∞

0

r|u^2 |dr= 2 k

∫∞

0

re−^2 krdr

=

1

2 k

=

1

2 × 0. 232

= 2 .16 fm

3.22 The inside wave functionu 1 =Asinkris maximum atr≈R. Therefore
kR=π/ 2

or k^2 R^2 =M(V 0 −W)R^2 /^2 =π^2 / 4

V 0 =

π^2 ^2 c^2

4 Mc^2 R^2

+W

Substitutingc= 197 .3MeV−fm, Mc^2 =940 MeV, R = 1 .5 fm and

W= 2 .2 MeV, we findV 0 ≈47 MeV

3.23 <r^2 >=

∫

ψ∗r^2 ψdτ

=

∫∞

0

r^2

r^2

(α

2 π

)

e−^2 αr 4 πr^2 dr

=

1

2 α^2

√

<r^2 >=

1

√

2 α

=

4. 3 × 10 −^15 m

√

2

= 3. 0 × 10 −^15 m= 3 .0fm

3.24 Referring to Problem 3.18, the energy of thenth level is

En=

n^2 h^2

8 mL^2

(1)

and

En+ 1 =

(n+1)^2 h^2

8 ML^2

(2)

ThereforeEn+ 1 −En=

(2n+1)h^2

8 mL^2

(3)